Question 315269
a. Equation is
 
{{{2x^2 + 5y^2 = 4}}}

Or, {{{y=sqrt(4-2x^2)/5}}}, where, {{{ -sqrt(2) <= x <= sqrt(2)}}}  and {{{ 0 <= y <= 2/5)}}}  

Hence, equation of the inverse relation is

{{{x=sqrt(4-2y^2)/5}}}, where, {{{ -sqrt(2) <= y <= sqrt(2)}}}  and {{{ 0 <= x <= 2/5)}}}  

Or,{{{2y^2 + 5x^2 = 4}}}where, {{{ -sqrt(2) <= y <= sqrt(2)}}}  and {{{ 0 <= x <= 2/5)}}}  

b. Equation is

{{{y=3x^2-5x+9}}}where, {{{ y >0}}}  for any x

Hence, equation of the inverse relation is

{{{x=3y^2-5y+9}}}where, {{{ x >0}}}  for any y